# Finding the graph formula with know points and other equation.

1. ### Pharrahnox

106
I have an equation for determining the acceleration of an object being propelled by a constant power source, that is affected by air resistance:

a = $\frac{P}{mv}$-$\frac{CDpAv2}{2m}$

Since F = $\frac{P}{v}$

I am trying to graph this as a velocity-time graph, however, I don't know how to do it. There is no time variable that I can replace with x, and the y-value (velocity) is mixed into the equation already.

I remember the equation given to me for a similar sort of thing, without air resitance, but instead just a constant friction force, that was something like this:

y = k(1-e-ax)

Where k is a constant, which is the maximum speed, and a is another constant which represents the force of air resistance.

The maximum speed in this case is $\sqrt[3]{\frac{2P}{CDpA}}$, so the equation would be something like:

y = $\sqrt[3]{\frac{2P}{CDpA}}$(1-e-ax)

But that's as far as I've gotten. By the use of iteration, I have determined the velocity at several different times, here's a few, just in case it helps:

(0,0) (25,83.4762) (50,118.1195) (75,126.1601) (100,127.7624) (125,128.0709)

This is for variables of values: p = 0.001, A = 1900.933, m = 1900.933, P = 1*106
and CD = 0.5

Any help would be greatly appreciated, and if you need any more information, just let me know.

EDIT: the equations don't seem to be formatting correctly, so I'll redo them down here:

a = P/v - (Cd*p*A*v^2) /2
F = P/v
max speed = ( (2*P) / (Cd*p*A) )^1/3
y = ( (2*P) / (Cd*p*A) )^1/3 * (1 - e^-ax)

Last edited: Mar 11, 2014
2. ### gopher_p

575
Given ##a=\frac{dv}{dt}##, what you have is a separable ordinary differential equation $$\frac{dv}{dt}=\frac{\alpha}{v}-\beta v^2$$ where ##\alpha=\frac{P}{m}## and ##\beta=\frac{C_DpA}{2m}## have been substituted to make the equation (slightly) more manageable. Hopefully that's enough to get you going.

I'm not sure where you got your force equation. Are you not using ##F=ma##? Furthermore, I don't know that it's even useful here.

I reckon it's doubtful that the equation that works in a constant friction situation applies here. So you're likely barking up the wrong tree there.

3. ### Pharrahnox

106
I have never done differential equations before, and I tried to just find the anti-derivative, but my teacher said that I had done it incorrectly, as I differentiated dv instead of dt, or the other way around. Anyway, here is what I got:

v = (Pt/m)*ln(|v|)-(CDpAv3t)/(6m)+c

Unfortunately from there, if it is correct, I don't know where to go - how to get v by itself without v on the other side.

What do I do from here, or from the start if that isn't correct?

Last edited: Mar 12, 2014
4. ### Stephen Tashi

4,413
I think this is due to your using the SUB and SUP tags instead of the usual LaTex notation.

$a = \frac{P}{v} - \frac{ (C_d p A v^2) }{2}$

5. ### Pharrahnox

106
Oh ok, I'll give it a go:

a = ($\frac{Pt}{m}$)ln(|v|)-$\frac{C_{D}pAv^{3}t}{6m}$

Seems to work, thanks.

6. ### Pharrahnox

106
Does anyone have any advice on what I should do to get time into the equation, without velocity on both sides? I don't even know what to type into google to find information on it. Any information would be much appreciated.